Question

p(k successes)=_nc_kp^k(1 - p)^{n - k}
_nc_k=\frac{n!}{(n - k)!k!}
○ _7c_5(\frac{1}{6})^2(\frac{1}{6})^5
○ _7c_5(\frac{1}{6})^5(\frac{5}{6})^2
○ _7c_2(\frac{1}{6})^2(\frac{5}{6})^5
○ _7c_2(\frac{2}{6})^2(\frac{4}{6})^5

Explanation:

Step1: Recall binomial - probability formula

The binomial - probability formula is $P(k\text{ successes})={}_{n}C_{k}p^{k}(1 - p)^{n - k}$, where ${}_{n}C_{k}=\frac{n!}{(n - k)!k!}$, $n$ is the number of trials, $k$ is the number of successes, and $p$ is the probability of success in a single trial.

Step2: Analyze the general form

We assume that in a binomial - experiment, if the probability of success in a single trial is $p$ and the probability of failure is $1 - p$, and we have $n$ trials with $k$ successes.
Let's assume a fair - die rolling experiment. If we roll a fair six - sided die, the probability of getting a particular number (say, 1) in a single roll is $p=\frac{1}{6}$, and the probability of not getting that number is $1 - p=\frac{5}{6}$. If we roll the die $n = 7$ times and we want to find the probability of getting $k = 2$ successes (getting the particular number 2 times).
Using the binomial - probability formula $P(k\text{ successes})={}_{n}C_{k}p^{k}(1 - p)^{n - k}$, with $n = 7$, $k = 2$, and $p=\frac{1}{6}$, we get $P(2\text{ successes})={}_{7}C_{2}(\frac{1}{6})^{2}(\frac{5}{6})^{5}$.

Answer:

$_{7}C_{2}(\frac{1}{6})^{2}(\frac{5}{6})^{5}$ (corresponds to the third option)